= Solution
In the <classical risk model>, let $N(t)$ be the <Poisson process> of claim arrivals, independent of the independent claim sizes. The reserve starts at $u$ and earns premiums continuously at rate $c$, so
$$
\boxed{U(t)=u+ct-\sum_{i=1}^{N(t)}X_i,\qquad\psi(u)=\mathbb P_u(\tau<\infty),\quad\tau=\inf\{t\ge0:U(t)<0\}.}
$$
Here $\psi$ is the <ultimate ruin probability> and $\phi=1-\psi$ is the <survival probability>. The <relative safety loading> is $\theta=(c-\lambda\mu)/(\lambda\mu)>0$.
To derive the <survival integro-differential equation>, condition on the first claim time $T$, which has the <exponential distribution> of rate $\lambda$. Immediately before that claim the available capital is $u+cT$. A claim $x\le u+cT$ leaves survival probability $\phi(u+cT-x)$, since the future claim process restarts with the same independent law; a larger claim causes ruin immediately. The <first-claim decomposition for survival probability> is therefore
$$
\phi(u)=\int_0^\infty\lambda e^{-\lambda t}\int_0^{u+ct}\phi(u+ct-x)f(x)\,dx\,dt.
$$
Put $r=\lambda/c$ and $H(v)=\int_0^v\phi(v-x)f(x)dx$. Changing variables to $v=u+ct$ yields
$$
\phi(u)=r e^{ru}\int_u^\infty e^{-rv}H(v)\,dv.
$$
This representation justifies differentiation. The function $H$ is continuous: extend $\phi$ by zero to negative arguments, then use continuity of translations of the integrable density $f$ in $L^1$ to control its convolution with the bounded function $\phi$. Differentiating the lower endpoint of the displayed integral gives $\phi'=r\phi-rH$. Since $c=(1+\theta)\lambda\mu$,
$$
\boxed{\phi'(u)=\frac{\phi(u)}{(1+\theta)\mu}-\frac1{(1+\theta)\mu}\int_0^u\phi(u-x)f(x)\,dx.}
$$
This deduction uses the first-arrival law rather than taking the desired differential equation as an assumption.
For $f(x)=4xe^{-2x}$, the claim law is an <Erlang distribution> of shape two and rate two. Its mean is $\mu=4\int_0^\infty x^2e^{-2x}dx=1$. With $\theta=2$, we have $r=1/3$. Changing the convolution variable to $t=u-x$ gives
$$
H(u)=4e^{-2u}\int_0^u(u-t)\phi(t)e^{2t}dt=4e^{-2u}I(u).
$$
Hence \b[the first requested specialization] is
$$
\boxed{3\phi'(u)=\phi(u)-4e^{-2u}I(u).}
$$
Let $J(u)=\int_0^u\phi(t)e^{2t}dt$. Differentiation under the finite integral gives $I'=J$ and $J'=\phi e^{2u}$. Differentiating the preceding equation and using $4e^{-2u}I=\phi-3\phi'$ then gives
$$
3\phi''=\phi'+8e^{-2u}I-4e^{-2u}J=-5\phi'+2\phi-4e^{-2u}J.
$$
Thus \b[the second identity] is
$$
\boxed{3\phi''(u)=-5\phi'(u)+2\phi(u)-4e^{-2u}\int_0^u\phi(t)e^{2t}dt.}
$$
Differentiate once more:
$$
3\phi'''=-5\phi''+2\phi'+8e^{-2u}J-4\phi.
$$
The preceding second-derivative equation gives $8e^{-2u}J=-6\phi''-10\phi'+4\phi$. Substitution cancels the undifferentiated terms and proves
$$
\boxed{3\phi'''+11\phi''+8\phi'=0.}
$$
This is the <Erlang claim-size differential equation for survival probability> with $\beta=2$ and $r=1/3$.
Its characteristic polynomial factors as $z(3z+8)(z+1)$. The limit $\phi(u)\to1$ therefore makes the general bounded solution
$$
\phi(u)=1+A e^{-u}+B e^{-8u/3}.
$$
The supplied <zero-capital survival probability> is $\phi(0)=2/3$, so $A+B=-1/3$. We also need the initial derivative carried by the original integral equation: $I(0)=0$ implies $\phi'(0)=\phi(0)/3=2/9$, giving $-A-(8/3)B=2/9$. Solving gives $A=-2/5$ and $B=1/15$. Therefore \b[the survival and ruin probabilities are]
$$
\boxed{\phi(u)=1-\frac25e^{-u}+\frac1{15}e^{-8u/3},\qquad\psi(u)=\frac25e^{-u}-\frac1{15}e^{-8u/3}\quad(u\ge0).}
$$
The extra initial derivative is essential: $\phi(0)$ and the limit at infinity alone would leave one free parameter in the third-order equation. The result has $\psi(0)=1/3$ and tends to zero. It is positive and decreasing, since $\psi'(u)=-(2/5)e^{-u}+(8/45)e^{-8u/3}<0$ for $u\ge0$. These signs agree with its interpretation as an <ultimate ruin probability>.
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